Series question

1. The problem statement, all variables and given/known data
[tex]a_{n} -> a [/tex] , a>0 and [tex]a_{n}>0[/tex] for all n. Prove that if [tex]b_{n} -> b[/tex] then [tex]{(a_{n})}^{(b_{n})} -> a^b[/tex].
(-> means the limit as n goes to infinity).

2. Relevant equations

3. The attempt at a solution

I first split up a_n into to sub series: [tex]a_{n_{k}}[/tex] and [tex]a_{n_{j}}[/tex] where for all k [tex]a_{n_{k}} >= 1 [/tex]and for all j [tex]0<a_{n_{j}}<1[/tex] . Now, starting with the first series, for all E>0 we can find a number N so that for all k>N [tex] b-E < b_{n_{k}} < b+E [/tex]and so
[tex]{(a_{n_{k}})}^{(b-E)} < {(a_{n_{k}})}^{(b_{n_{k}})} < {(a_{n_{k}})}^{(b+E)} [/tex]. I know that [tex] {(a_{n_{k}})}^{(b+E)} -> a^{(b+E)} [/tex] and so that since E can be as small as we want we should be able to use the sandwich rule to prove that [tex] {(a_{n_{k}})}^{(b_{n_{k}})} -> a^b [/tex] and we can do the same to prove that [tex] {(a_{n_{j}})}^{(b_{n_{j}})} -> a^b [/tex] . But how can I write it in a rigorous way?
Thanks.

If the limit as n goes to infinity of (a_n)^(b_n) is a^b, then for any E > 0, there exists an N such that for all indices i >= N, |(a_i)^(b_i) - a^b| < E. To demonstrate this (rigorously), you will need to tell me what this N is.

Well, if you don't mind a cheat-ish answer, you could just look at the logarithm, then use the facts that logarithmic and exponential functions are continuous and that products of convergent sequences converge to the product of the limits

What I was thinking is the following: Let N be the greater of [itex]N_a[/itex] and [itex]N_b[/itex]. Then surely [itex](a - E)^{b - E} < a_i^{b_i} < (a + E)^{b + E}[/itex]. Of course, what I really need to show is that [itex]a^b - E < a_i^{b_i} < a^b + E[/itex]. This is where I'm stumped.

Thanks, is this what you meant:
for all m [tex]{a_m}^{b_n} -> {a_m}^b[/tex]so for every E there's some N so that for all n>N [tex]{a_m}^b -E < {a_m}^{b_n} < {a_m}^b + E[/tex] and since [tex]{a_m}^{b} -> {a}^b[/tex]so for every F there's some M so that for all m>M [tex]{a}^b -F < {a_m}^{b} < {a}^b + F[/tex] and so if P>max{M,N} then for all m,n > P
[tex]{a_m}^b -(E+F) < {a_m}^{b_n} < {a}^b + (E+F)[/tex] QED
Is that right?
Thanks.